Proving the Hyperbolic Function Identity (1+tanhx)/(1-tanhx)=e^(2x)

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SUMMARY

The identity (1+tanhx)/(1-tanhx)=e^(2x) is proven by substituting tanhx with sinhx/coshx. The simplification process involves rewriting the expression as (1 + sinhx/coshx)/(1 - sinhx/coshx) and eliminating internal fractions to arrive at (coshx+sinhx)/(coshx-sinhx). This ultimately simplifies to e^(2x), confirming the identity as established fact.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically tanh, sinh, and cosh.
  • Familiarity with algebraic manipulation of fractions.
  • Knowledge of exponential functions and their properties.
  • Basic calculus concepts related to limits and continuity (optional for deeper understanding).
NEXT STEPS
  • Study hyperbolic function identities and their proofs.
  • Learn about the properties of exponential functions in greater detail.
  • Explore algebraic techniques for simplifying complex fractions.
  • Investigate the applications of hyperbolic functions in calculus and differential equations.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding hyperbolic functions and their identities.

Pietair
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Homework Statement


Prove that:
(1+tanhx)/(1-tanhx)=e^(2x)

Homework Equations



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The Attempt at a Solution



I tried substituting tanhx for (e^x-e^(-x))/(e^x+e^(-x)) and for (e^(2x)-1)/(e^(2x)+1))

But I really have no clue how to continue...
 
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Hi Pietair! :smile:

Hint : (1+tanhx)/(1-tanhx) = (1 + sinhx/coshx)/(1 - sinhx/coshx) = … ? :wink:
 
Thanks for your answer but it still doesn't make sense.

I don't know how to rewrite it to something more "common".
 
Pietair said:
Thanks for your answer but it still doesn't make sense.

I don't know how to rewrite it to something more "common".

try simplifying (1 + sinhx/coshx)/(1 - sinhx/coshx) …

get rid of the internal fractions :wink:
 
(coshx+sinhx)/(coshx-sinhx)

= (0.5e^x+0.5e^(-x)+0.5e^x-0.5e^(-x))/((0.5e^x+0.5e^(-x)-0.5e^x+0.5e^(-x))

= e^x/e^(-x)

= e^2x (proven)

Thanks a lot!
 

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